zoukankan      html  css  js  c++  java
  • Hdu 1017 A Mathematical Curiosity

     A Mathematical Curiosity

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 41387    Accepted Submission(s): 13311

    Problem Description

    Given two integers n and m, count the number of pairs of integers (a,b) such that 0 < a < b < n and (a^2+b^2 +m)/(ab) is an integer.

    This problem contains multiple test cases!

    The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

    The output format consists of N output blocks. There is a blank line between output blocks.

    Input

    You will be given a number of cases in the input. Each case is specified by a line containing the integers n and m. The end of input is indicated by a case in which n = m = 0. You may assume that 0 < n <= 100.

    Output

    For each case, print the case number as well as the number of pairs (a,b) satisfying the given property. Print the output for each case on one line in the format as shown below.

    Sample Input

    1

    10 1

    20 3

    30 4

    0 0

    Sample Output

    Case 1: 2

    Case 2: 4

    Case 3: 5

    #include<stdio.h>int main()
    {
        int i;
        int N;
        int n,m;
        int a,b;
        int count,x;
    
        scanf("%d",&N);
        for(i=0;i<N;i++)
        {
            x=1;
            while(scanf("%d %d",&n,&m),n>0)
            {
                count=0;
                for(a=1;a<n-1;a++)
                    for(b=a+1;b<n;b++)
                        if((a*a+b*b+m)%(a*b)==0)
                            count++;
                printf("Case %d: %d
    ",x++,count);
            }
            if(i!=N-1)
                printf("
    ");
        }
        return 0;
    }
    

      

  • 相关阅读:
    zookeeper 分布式锁
    mysql linux 安装
    分布式配置中心Apollo
    分布式任务调度平台xxl-job
    Java并发编程笔记之ThreadLocalRandom源码分析
    Java并发编程笔记之ThreadLocal源码分析
    SpringCloud实战10-Sleuth
    SpringCloud实战9-Stream消息驱动
    SpringCloud实战8-Bus消息总线
    SpringCloud实战7-Config分布式配置管理
  • 原文地址:https://www.cnblogs.com/zhangliu/p/7057789.html
Copyright © 2011-2022 走看看